Problem: Suppose we have a vector field $f(x, y) = \left( -y, x \right)$ and a curve $C$ that is parameterized by $\alpha(t) = (\cos(t), \sin(t))$ for $0 < t < 2\pi$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Answer: Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = \left( -y, x \right)$ and $\alpha(t) = (\cos(t), \sin(t))$. $\begin{aligned} &f(\alpha(t)) = (-\sin(t), \cos(t)) \\ \\ &\alpha'(t) = (-\sin(t), \cos(t)) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_0^{2\pi} (-\sin(t), \cos(t)) \cdot (-\sin(t), \cos(t)) \, dt$ Let's solve the integral. $\begin{aligned} &\int_0^{2\pi} (-\sin(t), \cos(t)) \cdot (-\sin(t), \cos(t)) \, dt \\ \\ &= \int_0^{2\pi} \sin^2(t) + \cos^2(t) \, dt \\ \\ &= \int_0^{2\pi} 1 \, dt \\ \\ &= 2\pi \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = 2\pi$.